\(\int \frac {1}{x^4 (a+b x^{3/2})^{2/3}} \, dx\) [2280]

Optimal result

Integrand size = 17, antiderivative size = 148 \[ \int \frac {1}{x^4 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {\sqrt [3]{a+b x^{3/2}}}{3 a x^3}+\frac {5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}-\frac {10 b^2 \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^{3/2}}}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{8/3}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{8/3}} \]

[Out]

-1/3*(a+b*x^(3/2))^(1/3)/a/x^3+5/9*b*(a+b*x^(3/2))^(1/3)/a^2/x^(3/2)-5/18*b^2*ln(x)/a^(8/3)+5/9*b^2*ln(a^(1/3)
-(a+b*x^(3/2))^(1/3))/a^(8/3)-10/27*b^2*arctan(1/3*(a^(1/3)+2*(a+b*x^(3/2))^(1/3))/a^(1/3)*3^(1/2))/a^(8/3)*3^
(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {272, 44, 59, 631, 210, 31} \[ \int \frac {1}{x^4 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {10 b^2 \arctan \left (\frac {2 \sqrt [3]{a+b x^{3/2}}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{8/3}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}-\frac {\sqrt [3]{a+b x^{3/2}}}{3 a x^3} \]

[In]

Int[1/(x^4*(a + b*x^(3/2))^(2/3)),x]

[Out]

-1/3*(a + b*x^(3/2))^(1/3)/(a*x^3) + (5*b*(a + b*x^(3/2))^(1/3))/(9*a^2*x^(3/2)) - (10*b^2*ArcTan[(a^(1/3) + 2
*(a + b*x^(3/2))^(1/3))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(8/3)) - (5*b^2*Log[x])/(18*a^(8/3)) + (5*b^2*Log[a^(
1/3) - (a + b*x^(3/2))^(1/3)])/(9*a^(8/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 59

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} \text {Subst}\left (\int \frac {1}{x^3 (a+b x)^{2/3}} \, dx,x,x^{3/2}\right ) \\ & = -\frac {\sqrt [3]{a+b x^{3/2}}}{3 a x^3}-\frac {(5 b) \text {Subst}\left (\int \frac {1}{x^2 (a+b x)^{2/3}} \, dx,x,x^{3/2}\right )}{9 a} \\ & = -\frac {\sqrt [3]{a+b x^{3/2}}}{3 a x^3}+\frac {5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}+\frac {\left (10 b^2\right ) \text {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^{3/2}\right )}{27 a^2} \\ & = -\frac {\sqrt [3]{a+b x^{3/2}}}{3 a x^3}+\frac {5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{8/3}}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{7/3}} \\ & = -\frac {\sqrt [3]{a+b x^{3/2}}}{3 a x^3}+\frac {5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{8/3}}+\frac {\left (10 b^2\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^{3/2}}}{\sqrt [3]{a}}\right )}{9 a^{8/3}} \\ & = -\frac {\sqrt [3]{a+b x^{3/2}}}{3 a x^3}+\frac {5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}-\frac {10 b^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^{3/2}}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{9 \sqrt {3} a^{8/3}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{8/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^4 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {\frac {3 a^{2/3} \sqrt [3]{a+b x^{3/2}} \left (-3 a+5 b x^{3/2}\right )}{x^3}-10 \sqrt {3} b^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^{3/2}}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+10 b^2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^{3/2}}\right )-5 b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^{3/2}}+\left (a+b x^{3/2}\right )^{2/3}\right )}{27 a^{8/3}} \]

[In]

Integrate[1/(x^4*(a + b*x^(3/2))^(2/3)),x]

[Out]

((3*a^(2/3)*(a + b*x^(3/2))^(1/3)*(-3*a + 5*b*x^(3/2)))/x^3 - 10*Sqrt[3]*b^2*ArcTan[(1 + (2*(a + b*x^(3/2))^(1
/3))/a^(1/3))/Sqrt[3]] + 10*b^2*Log[-a^(1/3) + (a + b*x^(3/2))^(1/3)] - 5*b^2*Log[a^(2/3) + a^(1/3)*(a + b*x^(
3/2))^(1/3) + (a + b*x^(3/2))^(2/3)])/(27*a^(8/3))

Maple [F]

\[\int \frac {1}{x^{4} \left (a +b \,x^{\frac {3}{2}}\right )^{\frac {2}{3}}}d x\]

[In]

int(1/x^4/(a+b*x^(3/2))^(2/3),x)

[Out]

int(1/x^4/(a+b*x^(3/2))^(2/3),x)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\text {Timed out} \]

[In]

integrate(1/x^4/(a+b*x^(3/2))^(2/3),x, algorithm="fricas")

[Out]

Timed out

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.28 \[ \int \frac {1}{x^4 \left (a+b x^{3/2}\right )^{2/3}} \, dx=- \frac {2 \Gamma \left (\frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {8}{3} \\ \frac {11}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{\frac {3}{2}}}} \right )}}{3 b^{\frac {2}{3}} x^{4} \Gamma \left (\frac {11}{3}\right )} \]

[In]

integrate(1/x**4/(a+b*x**(3/2))**(2/3),x)

[Out]

-2*gamma(8/3)*hyper((2/3, 8/3), (11/3,), a*exp_polar(I*pi)/(b*x**(3/2)))/(3*b**(2/3)*x**4*gamma(11/3))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^4 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {10 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{27 \, a^{\frac {8}{3}}} - \frac {5 \, b^{2} \log \left ({\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}} + {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{27 \, a^{\frac {8}{3}}} + \frac {10 \, b^{2} \log \left ({\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{27 \, a^{\frac {8}{3}}} + \frac {5 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {4}{3}} b^{2} - 8 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a b^{2}}{9 \, {\left ({\left (b x^{\frac {3}{2}} + a\right )}^{2} a^{2} - 2 \, {\left (b x^{\frac {3}{2}} + a\right )} a^{3} + a^{4}\right )}} \]

[In]

integrate(1/x^4/(a+b*x^(3/2))^(2/3),x, algorithm="maxima")

[Out]

-10/27*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^(3/2) + a)^(1/3) + a^(1/3))/a^(1/3))/a^(8/3) - 5/27*b^2*log((b*x
^(3/2) + a)^(2/3) + (b*x^(3/2) + a)^(1/3)*a^(1/3) + a^(2/3))/a^(8/3) + 10/27*b^2*log((b*x^(3/2) + a)^(1/3) - a
^(1/3))/a^(8/3) + 1/9*(5*(b*x^(3/2) + a)^(4/3)*b^2 - 8*(b*x^(3/2) + a)^(1/3)*a*b^2)/((b*x^(3/2) + a)^2*a^2 - 2
*(b*x^(3/2) + a)*a^3 + a^4)

Giac [A] (verification not implemented)

none

Time = 1.55 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^4 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {\frac {10 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {8}{3}}} + \frac {5 \, b^{3} \log \left ({\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}} + {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {8}{3}}} - \frac {10 \, b^{3} \log \left ({\left | {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {8}{3}}} - \frac {3 \, {\left (5 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {4}{3}} b^{3} - 8 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a b^{3}\right )}}{a^{2} b^{2} x^{3}}}{27 \, b} \]

[In]

integrate(1/x^4/(a+b*x^(3/2))^(2/3),x, algorithm="giac")

[Out]

-1/27*(10*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x^(3/2) + a)^(1/3) + a^(1/3))/a^(1/3))/a^(8/3) + 5*b^3*log((b*x
^(3/2) + a)^(2/3) + (b*x^(3/2) + a)^(1/3)*a^(1/3) + a^(2/3))/a^(8/3) - 10*b^3*log(abs((b*x^(3/2) + a)^(1/3) -
a^(1/3)))/a^(8/3) - 3*(5*(b*x^(3/2) + a)^(4/3)*b^3 - 8*(b*x^(3/2) + a)^(1/3)*a*b^3)/(a^2*b^2*x^3))/b

Mupad [B] (verification not implemented)

Time = 6.18 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.39 \[ \int \frac {1}{x^4 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {10\,b^2\,\ln \left ({\left (a+b\,x^{3/2}\right )}^{1/3}-a^{1/3}\right )}{27\,a^{8/3}}-\frac {\frac {8\,b^2\,{\left (a+b\,x^{3/2}\right )}^{1/3}}{9\,a}-\frac {5\,b^2\,{\left (a+b\,x^{3/2}\right )}^{4/3}}{9\,a^2}}{{\left (a+b\,x^{3/2}\right )}^2-2\,a\,\left (a+b\,x^{3/2}\right )+a^2}+\frac {\ln \left (\frac {-5\,b^2+\sqrt {3}\,b^2\,5{}\mathrm {i}}{3\,a^{5/3}}-\frac {10\,b^2\,{\left (a+b\,x^{3/2}\right )}^{1/3}}{3\,a^2}\right )\,\left (-5\,b^2+\sqrt {3}\,b^2\,5{}\mathrm {i}\right )}{27\,a^{8/3}}-\frac {\ln \left (\frac {5\,b^2+\sqrt {3}\,b^2\,5{}\mathrm {i}}{3\,a^{5/3}}+\frac {10\,b^2\,{\left (a+b\,x^{3/2}\right )}^{1/3}}{3\,a^2}\right )\,\left (5\,b^2+\sqrt {3}\,b^2\,5{}\mathrm {i}\right )}{27\,a^{8/3}} \]

[In]

int(1/(x^4*(a + b*x^(3/2))^(2/3)),x)

[Out]

(10*b^2*log((a + b*x^(3/2))^(1/3) - a^(1/3)))/(27*a^(8/3)) - ((8*b^2*(a + b*x^(3/2))^(1/3))/(9*a) - (5*b^2*(a
+ b*x^(3/2))^(4/3))/(9*a^2))/((a + b*x^(3/2))^2 - 2*a*(a + b*x^(3/2)) + a^2) + (log((3^(1/2)*b^2*5i - 5*b^2)/(
3*a^(5/3)) - (10*b^2*(a + b*x^(3/2))^(1/3))/(3*a^2))*(3^(1/2)*b^2*5i - 5*b^2))/(27*a^(8/3)) - (log((3^(1/2)*b^
2*5i + 5*b^2)/(3*a^(5/3)) + (10*b^2*(a + b*x^(3/2))^(1/3))/(3*a^2))*(3^(1/2)*b^2*5i + 5*b^2))/(27*a^(8/3))